Question

The function f(t)=3t2.

Answer 1

Answer:

Since, for a function f(x),

$\text{Average rate of change between }x_1\text{ and }x_2=\frac{f(x_2)-f(x_1)}{x_2-x_1}$

Given function,

$f(t)=3t^2$

Thus,

(a) the average rate of change of f(t) between t=1 and t=1.001

$=\frac{f(1.001)-f(1)}{1.001-1}$

$=\frac{3(1.001)^2-3(1)^2}{0.001}$

$=6.003$

(b) the average rate of change of f(t) between t=2 and t=2.001

$=\frac{f(2.001)-f(2)}{2.001-2}$

$=\frac{3(2.001)^2-3(2)^2}{0.001}$

$=12.003$

(c) the average rate of change of f(t) between t=3 and t=3.001

$=\frac{f(3.001)-f(3)}{3.001-3}$

$=\frac{3(3.001)^2-3(3)^2}{0.001}$

$=18.003$

(d) the average rate of change of f(t) between t=4 and t=4.001

$=\frac{f(4.001)-f(4)}{4.001-4}$

$=\frac{3(4.001)^2-3(4)^2}{0.001}$

$=24.003$

$=6.003$

Answer 2

$\bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\ -------------------------------\\\\ f(t)= 3t^2 \qquad \begin{cases} t_1=1\\ t_2=1.001 \end{cases}\implies \cfrac{f(1.001)-f(1)}{1.001-1} \\\\\\ \cfrac{3(1.001)^2~~-~~3(1)^2}{0.001}\implies \cfrac{3.006003~-~3}{0.001} \\\\\\ \cfrac{0.006003}{0.001}\implies 6.003$